 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/8a/d/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

On the first and third coordinates, the identity

S J S = J

from part (b) gives

e^{x B} C e^{x B} = C

; the second coordinate is also unchanged. Thus

n = 1 \sum N (e^{x B} C e^{x B})^{n} = diag (N, \frac{( - 1 ) ^{N} - 1}{2}, \frac{( - 1 ) ^{N} - 1}{2}) .

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